Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.StringTokenizer;

public class Main {
	
	public static void main(String[] args) throws IOException {
		
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st;
		
		int N = Integer.parseInt(br.readLine());
		double[] p = new double[N];
		st = new StringTokenizer(br.readLine());
		for(int i=0; i<N; i++) {
			p[i] = Double.parseDouble(st.nextToken());
		}
		
		double[][] dp = new double[N+1][N+1];
		dp[0][0] = 1;
		dp[1][1] = p[0];
		for(int tosses=1; tosses<=N; tosses++) {
			dp[tosses][0] = dp[tosses-1][0] * (1-p[tosses-1]);
			for(int heads=1; heads<=N; heads++) {
				dp[tosses][heads] = dp[tosses-1][heads-1]*p[tosses-1] + (1-p[tosses-1])*dp[tosses-1][heads]; 
			}
		}
		
		double ans = 0;
		for(int i=N/2+1; i<=N; i++) {
			ans += dp[N][i];
		}
		System.out.println(ans);
		
		
		
		
	}
}